Show that $\int_{\pi/6}^{\pi/4} \cot x dx= 1/2 \ln 2$

Question

I took $\cot x$ as $\cos x$ divided by $\sin x$ . Substituted $u = \sin x$ , $dx = 1 / \cos x du$. Got $\ln u$ . Replaced $u = \sin x$ and put the limits in . Got $\ln \sqrt{ 2}$. What should I do next? .

Answer 1

We have $\ln(2^{1/2})=\frac{1}{2}\ln 2$. In general, if $a$ is positive, $\ln(a^x)=x\ln a$.

Remark: When we do the integration and substitute, the "raw" expression we get is $$\ln(1/\sqrt{2})-\ln(1/2).$$ There are various ways to simplify. Maybe most natural is to rewrite as $\ln(2)-\ln(\sqrt{2})$. Or else we can use the fact that $\ln(b)-\ln(a)=\ln(b/a)$.